Problem: Solve for $x$ and $y$ using substitution. ${x+3y = 7}$ ${x = 6y-2}$
Solution: Since $x$ has already been solved for, substitute $6y-2$ for $x$ in the first equation. ${(6y-2)}{+ 3y = 7}$ Simplify and solve for $y$ $6y-2 + 3y = 7$ $9y-2 = 7$ $9y-2{+2} = 7{+2}$ $9y = 9$ $\dfrac{9y}{{9}} = \dfrac{9}{{9}}$ ${y = 1}$ Now that you know ${y = 1}$ , plug it back into $\thinspace {x = 6y-2}\thinspace$ to find $x$ ${x = 6}{(1)}{ - 2}$ $x = 6 - 2$ ${x = 4}$ You can also plug ${y = 1}$ into $\thinspace {x+3y = 7}\thinspace$ and get the same answer for $x$ : ${x + 3}{(1)}{= 7}$ ${x = 4}$